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QveST [7]
3 years ago
10

B. The role of the moon is greater than that of the sun in the occurrence of tides. ???​

Physics
1 answer:
ahrayia [7]3 years ago
7 0

Our sun is 27 million times larger than our moon. Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth. If tidal forces were based solely on comparative masses, the sun should have a tide-generating force that is 27 million times greater than that of the moon. However, the sun is 390 times further from the Earth than is the moon. Thus, its tide-generating force is reduced by 3903, or about 59 million times less than the moon. Because of these conditions, the sun’s tide-generating force is about half that of the moon.

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A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to
irinina [24]

When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

<h3>Frictional force between the block and the horizontal surface</h3>

The frictional force between the block and the horizontal surface is determined by applying Newton's law;

∑F = ma

F - Ff = ma

Ff = F - ma

Ff = 4 - 2(1.2)

Ff = 4 - 2.4

Ff = 1.6 N

When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;

F - Ff = ma

5 - 1.6 = 2a

3.4 = 2a

a = 3.4/2

a = 1.7 m/s²

Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

Learn more about frictional force here: brainly.com/question/4618599

8 0
2 years ago
A home run is hit in such a way that the baseball just clears a wall 21m hgh located 130m from home plate the ball is hit at an
svlad2 [7]

Answer:

see the attachment

Explanation:

take coordinate system correctly. use formulas of projectile motion

Download pdf
3 0
2 years ago
suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance be
d1i1m1o1n [39]
Use the Inverse square law, Intensity (I) of a light is inversely proportional to the square of the distance(d).

I=1/(d*d)

Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.

L1/L2=(D2*D2)/(D1*D1)

L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>

7 0
2 years ago
When can a high speed velocity cause damage?'
sweet-ann [11.9K]

Answer:

50 Mph.

Explanation:

According to the National Severe Storms Laboratory, winds can really begin to cause damage when they reach <em><u>50 mph</u></em>. But here’s what happens before and after they reach that threshold, according to the Beaufort Wind Scale (showing estimated wind speeds): - at 19 to 24 mph, smaller trees begin to sway.

7 0
2 years ago
A lead nucleus is spherical with a radius of about 7 ✕ 10⁻¹⁵ m. The nucleus contains 82 protons (and typically 126 neutrons). Be
Komok [63]

Answer:

∈=3.1584x10^{26} \frac{V}{m}

Explanation:

Using the Gauss Law to determine the electric field of the net flux at the surface of the nucleus

∈=\frac{P}{E_{o}}

The P is the charge density and 'Eo' is the constant of permittivity in free space

to find P

P=\frac{q}{V}

V=\frac{4}{3}*\pi*r^3

V=\frac{4}{3}\pi*(7x10^{-15})^3

V=2.932x10^{-14} m^3

P=\frac{82C}{2.932x10^{-14}m^3}=2.7965x10^{15} \frac{C}{m^3}

So replacing

∈=\frac{2.7965x10^{15}\frac{C}{m^3}}{8.8542x10^{-12}\frac{C^2}{N*m^2}}

∈=3.1584x10^{26} \frac{V}{m}

3 0
3 years ago
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